Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int a[2][10514];
int b[2][10514];
int n;
typedef struct {
  int x, y;
} ele;
vector<ele> o[2];
void myswap(int k, int x, int y) {
  if (x == y) return;
  swap(a[k][x], a[k][y]);
  o[k].push_back((ele){x, y});
  o[k].push_back((ele){y, x});
  o[k].push_back((ele){x, y});
}
void myxxx(int k, int x, int y) {
  a[k][x] ^= a[k][y];
  o[k].push_back((ele){x, y});
}
int lb(int k) {
  if (k == 0) return 0;
  return k & -k;
}
void sol() {
  for (int i = 1; i <= n; i++) {
    scanf("%d", a[0] + i);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", a[1] + i);
  }
  memcpy(b, a, sizeof(a));
  for (int t = 0; t < 2; t++) {
    int flag;
    for (int k = 0, j = 1;;) {
      flag = 0;
      for (int i = j; i <= n; i++) {
        if (a[t][i] & (1 << k)) {
          if (i != j) {
            myswap(t, i, j);
          }
          flag = 1;
          break;
        }
      }
      if (flag == 0) {
        if (++k == 30) {
          break;
        }
        continue;
      }
      for (int i = 1; i <= n; i++) {
        if (i != j && (a[t][i] & (1 << k))) {
          myxxx(t, i, j);
        }
      }
      j++;
    }
  }
  for (int i = 1; a[1][i]; i++) {
    int j = i;
    for (; a[0][j] && lb(a[0][j]) != lb(a[1][i]); j++)
      ;
    if (a[0][j] == 0) break;
    myswap(0, i, j);
    int tmp = a[1][i] ^ a[0][i];
    for (int j = i + 1; a[0][j]; j++) {
      if (lb(a[0][j]) == lb(tmp)) {
        tmp ^= a[0][j];
        myxxx(0, i, j);
      }
    }
  }
  int flag = 1;
  int s;
  for (s = 1; a[1][s]; s++) {
    if (a[0][s] != a[1][s]) {
      flag = 0;
      break;
    }
  }
  for (; s <= n; s++) {
    myxxx(0, s, s);
  }
  for (int i = 0; i < (int)o[0].size(); i++) {
    int x = o[0][i].x;
    int y = o[0][i].y;
    b[0][x] ^= b[0][y];
  }
  for (int i = (int)o[1].size() - 1; i >= 0; i--) {
    int x = o[1][i].x;
    int y = o[1][i].y;
    b[0][x] ^= b[0][y];
  }
  for (int i = 1; i <= n; i++) {
    if (b[0][i] != b[1][i]) {
      flag = -1;
      break;
    }
  }
  if (flag == 1) {
    printf("%u\n", o[0].size() + o[1].size());
    for (int i = 0; i < (int)o[0].size(); i++) {
      printf("%d %d\n", o[0][i].x, o[0][i].y);
    }
    for (int i = (int)o[1].size() - 1; i >= 0; i--) {
      printf("%d %d\n", o[1][i].x, o[1][i].y);
    }
  } else {
    puts("-1");
  }
  o[0].clear();
  o[1].clear();
}
int main() {
  while (scanf("%d", &n) != EOF) sol();
  return 0;
}